(B) In the steady state,the heat flowing through rod $BC$ must equal the heat flowing through rod $CA$ because they are in series along the path $BCA$

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(B) In the steady state,the heat flowing through rod $BC$ must equal the heat flowing through rod $CA$ because they are in series along the path $BCA$.The rate of heat flow is given by $\frac{dQ}{dt} = \frac{kA(T_{high} - T_{low})}{L}$.For rod $BC$,length $L_{BC} = a$,temperature difference is $(\sqrt{2}T - T_C)$.For rod $CA$,length $L_{CA} = a\sqrt{2}$,temperature difference is $(T_C - T)$.Equating the rates of heat flow:$\frac{kA(\sqrt{2}T - T_C)}{a} = \frac{kA(T_C - T)}{a\sqrt{2}}$$\sqrt{2}(\sqrt{2}T - T_C) = T_C - T$$2T - \sqrt{2}T_C = T_C - T$$3T = T_C(1 + \sqrt{2})$$\frac{T_C}{T} = \frac{3}{1 + \sqrt{2}}$

Class 11 Physics 10-2.Heat Transfer Heat Conduction and Thermal Conductivity

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Three rods of identical cross-sectional area and made from the same metal form the sides of an isosceles triangle $ABC$,right-angled at $B$. The points $A$ and $B$ are maintained at temperatures $T$ and $\sqrt{2}T$ respectively. In the steady state,the temperature of point $C$ is $T_C$. Assuming that only heat conduction takes place,$\frac{T_C}{T}$ is equal to

IIT 1995 All IIT

A
$\frac{1}{(\sqrt{2} + 1)}$
B
$\frac{3}{(\sqrt{2} + 1)}$
C
$\frac{1}{2(\sqrt{2} - 1)}$
D
$\frac{1}{\sqrt{3}(\sqrt{2} - 1)}$

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10-2.Heat Transfer

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Heat Conduction and Thermal Conductivity

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The two ends of a rod of length $L$ and a uniform cross-sectional area $A$ are kept at two temperatures $T_{1}$ and $T_{2}$ $(T_{1} > T_{2})$. The rate of heat transfer,$\frac{dQ}{dt}$ through the rod in a steady state is given by:

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These days,people use steel utensils with copper bottoms. This is considered good for uniform heating of food. Explain this effect using the fact that copper is a better conductor of heat.

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Two plates of the same area are placed in contact. Their thicknesses as well as their thermal conductivities are in the ratio $2:3$. The outer surface of one plate is maintained at $10^{\circ} C$ and that of the other at $0^{\circ} C$. The temperature at the common surface is (in $^{\circ} C$)

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If $4 \, kg$ of ice is inside a closed cubical thermocol box of side length $20 \, cm$ and wall thickness $4 \, cm$, then the mass of the ice remaining after $10 \, hours$ is nearly. (Given: Outside temperature $= 50^{\circ}C$, Thermal conductivity of thermocol $K = 0.01 \, Js^{-1}m^{-1\circ}C^{-1}$, Latent heat of fusion of ice $L = 335 \times 10^3 \, Jkg^{-1}$) (in $ \, kg$)

MediumAP EAMCET 2017

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If the thermal conductivity of aluminum is $0.5 \ cal/cm \cdot s \cdot ^\circ C$,then the temperature gradient required to conduct $10 \ cal/s \cdot cm^2$ in the steady state is ...... $^\circ C/cm$.

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The two ends of a rod of length $L$ and a uniform cross-sectional area $A$ are kept at two temperatures $T_{1}$ and $T_{2}$ $(T_{1} > T_{2})$. The rate of heat transfer,$\frac{dQ}{dt}$ through the rod in a steady state is given by:

These days,people use steel utensils with copper bottoms. This is considered good for uniform heating of food. Explain this effect using the fact that copper is a better conductor of heat.

Two plates of the same area are placed in contact. Their thicknesses as well as their thermal conductivities are in the ratio $2:3$. The outer surface of one plate is maintained at $10^{\circ} C$ and that of the other at $0^{\circ} C$. The temperature at the common surface is (in $^{\circ} C$)

If the thermal conductivity of aluminum is $0.5 \ cal/cm \cdot s \cdot ^\circ C$,then the temperature gradient required to conduct $10 \ cal/s \cdot cm^2$ in the steady state is ...... $^\circ C/cm$.

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